The general iterative methods for equilibrium problems and fixed point problems of a countable family of nonexpansive mappings in Hilbert spaces

Abstract

In this paper, the researcher introduces the general iterative scheme for finding a common element of the set of equilibrium problems and fixed point problems of a countable family of nonexpansive mappings in Hilbert spaces. The results presented in this paper improve and extend the corresponding results announced by many others.

MSC:47H10, 47H09.

1 Introduction

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$. Let C be a nonempty closed and convex subset of H, and let $T:C\to C$ be a nonlinear mapping. In this paper, we use $F(T)$ to denote the fixed point set of T.

Recall the following definitions.

• (1) The mapping T is said to be nonexpansive if

  $$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

  (1.1)

Further, let F be a bifunction from $C\times C$ into ℝ, where ℝ is the set of real numbers. The so-called equilibrium problem for $F:C\times C\to \mathbb{R}$ is to find $y\in C$ such that

$$F(y,u)\ge 0,\mathrm{\forall }u\in C.$$

(1.2)

The set of solutions of (1.2) is denoted by $EP(F)$. Given a mapping $A:C\to H$, let $F(y,u)=〈Ay,u-y〉$ for all $y,u\in C$. Then $z\in EP(F)$ if and only if $〈Az,u-z〉\ge 0$ for all $u\in C$. Numerous problems in physics, optimization and economics reduce to finding a solution of (1.2).

• (2) The mappings ${\{T_n\}}_{n\in \mathbb{N}}$ are said to be a family of nonexpansive mappings from H into itself if

  $$\parallel T_{n}x-T_{n}y\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in H,$$

  (1.3)

and denoted by $F(T_n)=\{x\in H:T_{n}x=x\}$ is the fixed point set of $T_n$. Finding an optimal point in ${\bigcap }_{n\in \mathbb{N}}F(T_n)$ of the fixed point sets of each mapping is a matter of interest in various branches of science.

Recently, many authors considered the iterative methods for finding a common element of the set of solutions to problem (1.2) and of the set of fixed points of nonexpansive mappings; see, for example, [1, 2] and the references therein.

Next, let $A:C\to H$ be a nonlinear mapping. We recall the following definitions.

• (3) A is said to be monotone if

  $$〈Ax-Ay,x-y〉\ge 0,\mathrm{\forall }x,y\in C.$$

• (4) A is said to be strongly monotone if there exists a constant $\alpha >0$ such that

  $$〈Ax-Ay,x-y〉\ge \alpha {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

In such a case, A is said to be α-strongly monotone.

• (5) A is said to be inverse-strongly monotone if there exists a constant $\alpha >0$ such that

  $$〈Ax-Ay,x-y〉\ge \alpha {\parallel Ax-Ay\parallel }^2,\mathrm{\forall }x,y\in C.$$

In such a case, A is said to be α-inverse-strongly monotone.

The classical variational inequality is to find $u\in C$ such that

$$〈Au,v-u〉\ge 0,\mathrm{\forall }v\in C.$$

(1.4)

In this paper, we use $VI(C,A)$ to denote the set of solutions to problem (1.4). One can easily see that the variational inequality problem is equivalent to a fixed point problem. $u\in C$ is a solution to problem (1.4) if and only if u is a fixed point of the mapping $P_C(I-\lambda )T$, where $\lambda >0$ is a constant.

The variational inequality has been widely studied in the literature; see, for example, the work of Plubtieng and Punpaeng [3] and the references therein.

Recently, Ceng et al. [4] considered an iterative method for the system of variational inequalities (1.4). They got a strongly convergence theorem for problem (1.4) and a fixed point problem for a single nonexpansive mapping; see [4] for more details.

On the other hand, Moudafi [5] introduced the viscosity approximation method for nonexpansive mappings (see [6] for further developments in both Hilbert and Banach spaces).

A mapping $f:C\to C$ is called α-contractive if there exists a constant $\alpha \in (0,1)$ such that

$$\parallel f(x)-f(y)\parallel \le \alpha \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

(1.5)

Let f be a contraction on C. Starting with an arbitrary initial $x_1\in C$, define a sequence $\{x_n\}$ recursively by

$$x_{n+1}=(1-{\sigma }_n)T{x}_n+{\sigma }_{n}f(x_n),n\ge 0,$$

(1.6)

where $\{{\sigma }_n\}$ is a sequence in $(0,1)$. It is proved [5, 6] that under certain appropriate conditions imposed on $\{{\sigma }_n\}$, the sequence $\{x_n\}$ generated by (1.6) strongly converges to the unique solution q in C of the variational inequality

$$〈(I-f)q,p-q〉\ge 0,p\in C.$$

Let A be a strongly positive linear bounded operator on a Hilbert space H with a constant $\overline{\gamma }$; that is, there exists $\overline{\gamma }>0$ such that

$$〈Ax,x〉\ge \overline{\gamma }{\parallel x\parallel }^2,\mathrm{\forall }x\in H.$$

(1.7)

Recently, Marino and Xu [7] introduced the following general iterative method:

$$x_{n+1}=(I-{\alpha }_{n}A)T{x}_n+{\alpha }_n\gamma f(x_n),n\ge 0,$$

(1.8)

where A is a strongly positive bounded linear operator on H. They proved that if the sequence $\{{\alpha }_n\}$ of parameters satisfies appropriate conditions, then the sequence $\{x_n\}$ generated by (1.8) converges strongly to the unique solution of the variational inequality

$$〈(A-\gamma f)x^{\ast },x-x^{\ast }〉\ge 0,x\in C,$$

(1.9)

which is the optimality condition for the minimization problem

$$\underset{x\in C}{min}\frac{1}{2}〈Ax,x〉-h(x),$$

where h is a potential function for γf (i.e., $h^{\prime }(x)=\gamma f(x)$ for $x\in H$).

In 2007, Takahashi and Takahashi [2] introduced an iterative scheme by the viscosity approximation method for finding a common element of the set of solutions (1.2) and the set of fixed points of a nonexpansive mapping in Hilbert spaces. Let $S:C\to H$ be a nonexpansive mapping. Starting with arbitrary initial $x_1\in H$, define sequences $\{x_n\}$ and $\{u_n\}$ recursively by

$$\{\begin{array}{c}F(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\alpha }_n\gamma f(x_n)+(1-{\alpha }_n)S{u}_n,\mathrm{\forall }n\in \mathbb{N}.\hfill \end{array}$$

(1.10)

They proved that under certain appropriate conditions imposed on $\{{\alpha }_n\}$ and $\{r_n\}$, the sequences $\{x_n\}$ and $\{u_n\}$ converge strongly to $z\in F(S)\cap EP(F)$, where $z=P_{F(S)\cap EP(F)}f(z)$.

Next, Plubtieng and Punpaeng, [3] introduced an iterative scheme by the general iterative method for finding a common element of the set of solutions (1.2) and the set of fixed points of nonexpansive mappings in Hilbert spaces.

Let $S:H\to H$ be a nonexpansive mapping. Starting with an arbitrary $x_1\in H$, define sequences $\{x_n\}$ and $\{u_n\}$ by

$$\{\begin{array}{c}F(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)S{u}_n,\mathrm{\forall }n\in \mathbb{N}.\hfill \end{array}$$

(1.11)

They proved that if the sequences $\{{\alpha }_n\}$ and $\{r_n\}$ of parameters satisfy appropriate conditions, then the sequence $\{x_n\}$ generated by (1.11) converges strongly to the unique solution of the variational inequality

$$〈(A-\gamma f)z,x-z〉\ge 0,\mathrm{\forall }x\in F(S)\cap EP(F),$$

(1.12)

which is the optimality condition for the minimization problem

$$\underset{x\in F(S)\cap EP(F)}{min}\frac{1}{2}〈Ax,x〉-h(x),$$

where h is a potential function for γf (i.e., $h^{\prime }(x)=\gamma f(x)$ for $x\in H$).

Let $T_1,T_2,\dots$ be an infinite sequence of mappings of C into itself, and let ${\lambda }_1,{\lambda }_2,\dots$ be real numbers such that $0\le {\lambda }_i\le 1$ for every $i\in \mathbb{N}$. Then for any $n\in \mathbb{N}$, Takahashi [8] (see [9]) defined a mapping $W_n$ of C into itself as follows:

(1.13)

Such a mapping $W_n$ is called the W-mapping generated by $T_n,T_{n-1},\dots ,T_1$ and ${\lambda }_n,{\lambda }_{n-1},\dots ,{\lambda }_1$.

Recently, using process (1.13), Yao et al. [10] proved the following result.

Theorem 1.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $F:C\times C\to \mathbb{R}$ be an equilibrium bifunction satisfying the conditions:

• (1) F is monotone, that is, $F(x,y)+F(y,x)\le 0$ for all $x,y\in C$;

• (2) for each $x,y,z\in C$, ${lim}_{t\to 0}F(tz+(1-t)x,y)\le F(x,y)$;

• (3) for each $x\in C$, $y\mapsto F(x,y)$ is convex and lower semicontinuous.

Let ${\{T_i\}}_{i=1}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings of C into C such that ${\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)\cap EP(F)\ne \mathrm{\varnothing }$. Suppose $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are three sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n=1$ and $\{r_n\}\subset (0,\mathrm{\infty })$. Suppose the following conditions are satisfied:

• (1)
  $${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$$

  and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (2)
  $$0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$$

  ;

• (3)
  $${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$$

  and ${lim}_{n\to \mathrm{\infty }}(r_{n+1}-r_n)=0$.

Let f be a contraction of H into itself, and let $x_0\in H$ be given arbitrarily. Then the sequences $\{x_n\}$ and $\{y_n\}$ generated iteratively by

$$\{\begin{array}{c}F(y_n,x)+\frac{1}{r_n}〈x-y_n,y_n-x_n〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n{W}_n{y}_n,\hfill \end{array}$$

converge strongly to $x^{\ast }\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)\cap EP(F)$, the unique solution of the minimization problem

$$\underset{x\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)\cap EP(F)}{min}\frac{1}{2}{\parallel x\parallel }^2-h(x),$$

where h is a potential function for f.

Very recently, using process (1.13), Chen [11] proved the following result.

Theorem 1.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of nonexpansive mappings from C to C such that the common fixed point set $\mathrm{\Omega }={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)\ne \mathrm{\varnothing }$. Let $f:C\to H$ be an α-contraction, and let $A:H\to H$ be a self-adjoint, strongly positive bounded linear operator with a coefficient $\overline{\gamma }>0$. Let σ be a constant such that $0<\gamma \alpha <\overline{\gamma }$. For an arbitrary initial point $x_0$ belonging to C, one defines a sequence ${\{x_n\}}_{n\ge 0}$ iteratively

$$x_{n+1}=P_C[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{x}_n],\mathrm{\forall }n\ge 0,$$

(1.14)

where $\{{\alpha }_n\}$ is a real sequence in $[0,1]$. Assume the sequence $\{{\alpha }_n\}$ satisfies the following conditions:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$.

Then the sequence $\{x_n\}$ generated by (1.14) converges in norm to the unique solution $x^{\ast }$, which solves the following variational inequality:

$$x^{\ast }\in \mathrm{\Omega }\mathit{\text{such that}}〈(A-\gamma f)x^{\ast },x^{\ast }-\hat{x}〉\ge 0,\mathrm{\forall }\hat{x}\in \mathrm{\Omega }.$$

(1.15)

Motivated by this result, we introduce the following explicit general iterative scheme:

$$\{\begin{array}{c}{x}_1\in H,\hfill \\ F(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in H,\hfill \\ x_{n+1}=P_C[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n],\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

(1.16)

where ${\{T_n\}}_{n\in \mathbb{N}}$ is a family of nonexpansive mappings from H into itself such that ${\bigcap }_{n\in \mathbb{N}}F(T_n)$ is nonempty, $F:C\times C\to \mathbb{R}$ is an equilibrium bifunction, A is a strongly positive operator on H, f is a contraction of H into itself with $\alpha \in (0,1)$, $\{{\alpha }_n\}$, $\{r_n\}$, $\{{\lambda }_n\}$ suitable sequences in ℝ and $\{W_n\}$ is the sequence of a W-mapping generated by ${\{T_n\}}_{n\in \mathbb{N}}$ and $\{{\lambda }_n\}$. Let U be defined by $Ux={lim}_{n\to \mathrm{\infty }}{W}_{n}x={lim}_{n\to \mathrm{\infty }}{U}_{n,1}x$ for every $x\in C$ using process (1.13). We shall prove under mild conditions that $\{x_n\}$ and $\{u_n\}$ strongly converge to a point $x^{\ast }\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)\cap EP(F)$, which is the unique solution of the variational inequality

$$〈(A-\gamma f)x^{\ast },x^{\ast }-\hat{x}〉\ge 0,\mathrm{\forall }\hat{x}\in \bigcap _{i=1}^{\mathrm{\infty }}F(T_i)\cap EP(F),$$

(1.17)

or, equivalently, the unique solution of the minimization problem

$$\underset{x\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)\cap EP(F)}{min}\frac{1}{2}〈A\hat{x},\hat{x}〉-h(\hat{x}),$$

where h is a potential function for γf.

2 Preliminaries

Let H be a real Hilbert space with the norm $\parallel \cdot \parallel$ and the inner product $〈\cdot ,\cdot 〉$, and let C be a closed convex subset of H. We call $f:C\to H$ an α-contraction if there exists a constant $\alpha \in [0,1)$ such that

$$\parallel f(x)-f(y)\parallel \le \alpha \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

Let A be a strongly positive linear bounded operator on a Hilbert space H with a constant $\overline{\gamma }$; that is, there exists $\overline{\gamma }>0$ such that

$$〈Ax,x〉\ge \overline{\gamma }{\parallel x\parallel }^2,\mathrm{\forall }x\in H.$$

Next, we denote weak convergence and strong convergence by notations ⇀ and →, respectively. A space X is said to satisfy Opial’s condition [12] if for each sequence $\{x_n\}$ in X which converges weakly to a point $x\in X$, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-y\parallel ,\mathrm{\forall }y\in X,y\ne x.$$

For every point $x\in H$, there exists a unique nearest point in C, denoted by $P_{C}x$, such that

$$\parallel x-P_{C}x\parallel \le \parallel x-y\parallel ,\mathrm{\forall }y\in C.$$

$$P_C$$

is called the (nearest point or metric) projection of H onto C. In addition, $P_{C}x$ is characterized by the following properties: $P_{C}x\in C$ and

(2.1)

(2.2)

Recall that a mapping $T:H\to H$ is said to be firmly nonexpansive mapping if

$${\parallel Tx-Ty\parallel }^2\le 〈Tx-Ty,x-y〉,\mathrm{\forall }x,y\in H.$$

It is well known that $P_C$ is a firmly nonexpansive mapping of H onto C and satisfies

$${\parallel P_{C}x-P_{C}y\parallel }^2\le 〈x-y,P_{C}x-P_{C}y〉,\mathrm{\forall }x,y\in H.$$

(2.3)

If A is an α-inverse-strongly monotone mapping of C into H, then it is obvious that A is $\frac{1}{\alpha }$-Lipschitz continuous. We also have that for all $x,y\in C$ and $\lambda >0$,

$$\begin{array}{rcl}{\parallel (I-\lambda A)x-(I-\lambda A)y\parallel }^2& =& {\parallel x-y-\lambda (Ax-Ay)\parallel }^2\\ =& {\parallel x-y\parallel }^2-2\lambda 〈Ax-Ay,x-y〉+{\lambda }^2{\parallel Ax-Ay\parallel }^2\\ \le & {\parallel x-y\parallel }^2+\lambda (\lambda -2\alpha ){\parallel Ax-Ay\parallel }^2.\end{array}$$

(2.4)

So, if $\lambda \le 2\alpha$, then $I-\lambda A$ is a nonexpansive mapping of C into H.

The following lemmas will be useful for proving the convergence result of this paper.

Lemma 2.1 Let H be a real Hilbert space. Then for all $x,y\in H$,

• (1)
  $${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,x+y〉$$

  ;

• (2)
  $${\parallel x+y\parallel }^2\ge {\parallel x\parallel }^2+2〈y,x〉$$

  .

Lemma 2.2 ([13])

Let $\{x_n\}$ and $\{y_n\}$ be bounded sequences in a Banach space X, and let $\{{\beta }_n\}$ be a sequence in $[0,1]$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose that $x_{n+1}=(1-{\beta }_n)y_n+{\beta }_n{x}_n$ for all integers $n\ge 0$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. Then ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$.

Lemma 2.3 ([14])

Assume that $F:C\times C\to \mathbb{R}$, let us assume that F satisfies the following conditions:

(A1) $F(x,x)=0$ for all $x\in C$;

(A2) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$ for all $x,y\in C$;

(A3) for each $x,y,z\in C$, ${lim}_{t\to 0}F(tz+(1-t)x,y)\le F(x,y)$;

(A4) for each $x\in C$, $y\mapsto F(x,y)$ is convex and lower semicontinuous.

Lemma 2.4 ([14])

Assume that $F:C\times C\to \mathbb{R}$ satisfies (A1)-(A4). For $r>0$ and $x\in H$, define a mapping $T_r:H\to C$ as follows:

$$T_r(x)=\{z\in C:F(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\}$$

for all $z\in H$. Then the following hold:

• 1.
  $$T_r$$

  is single-valued;

• 2.
  $$T_r$$

  is firmly nonexpansive, i.e., for any $x,y\in H$,

  $${\parallel T_{r}x-T_{r}y\parallel }^2\le 〈T_{r}x-T_{r}y,x-y〉;$$
• 3.
  $$F(T_r)=EP(F)$$

  ;

• 4. EP(F) is closed and convex.

Lemma 2.5 ([12])

Let H be a Hilbert space, C be a closed convex subset of H, and $S:C\to C$ be a nonexpansive mapping with $F(S)\ne \mathrm{\varnothing }$. If $\{x_n\}$ is a sequence in C weakly converging to $x\in C$ and if $\{(I-S)x_n\}$ converges strongly to y, then $(I-S)x=y$.

Lemma 2.6 ([6])

Assume that $\{a_n\}$ is a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\delta }_n,n\ge 0,$$

where $\{{\gamma }_n\}$ is a sequence in $(0,1)$ and $\{{\delta }_n\}$ is a sequence in ℝ such that

• (1)
  $${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$$

  ;

• (2)
  $${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{{\delta }_n}{{\gamma }_n}\le 0$$

  or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_n|<\mathrm{\infty }$.

Then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

Lemma 2.7 ([7])

Let H be a Hilbert space, C be a nonempty closed convex subset of H, and $f:H\to H$ be a contraction with a coefficient $0<\alpha <1$, and let A be a strongly positive linear bounded operator with a coefficient $\overline{\gamma }>0$. Then, for $0<\gamma <\frac{\overline{\gamma }}{\alpha }$,

$$〈x-y,(A-\gamma f)x-(A-\gamma f)y〉\ge (\overline{\gamma }-\gamma \alpha ){\parallel x-y\parallel }^2,x,y\in H.$$

That is, $A-\gamma f$ is strongly monotone with a coefficient $\overline{\gamma }-\gamma \alpha$.

Lemma 2.8 ([7])

Assume A is a strongly positive linear bounded operator on a Hilbert space H with a coefficient $\overline{\gamma }>0$ and $0<\rho \le {\parallel A\parallel }^{-1}$. Then $\parallel I-\rho A\parallel \le 1-\rho \overline{\gamma }$.

Lemma 2.9 ([9] and [15])

Let C be a nonempty closed convex subset of a Banach space E. Let ${\{T_i\}}_{i=1}^{\mathrm{\infty }}$ be a sequence of nonexpansive mappings of C into itself with ${\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)\ne \mathrm{\varnothing }$, and let ${\{{\lambda }_i\}}_{i=1}^{\mathrm{\infty }}$ be a real sequence such that $0<{\lambda }_i\le b<1$, $\mathrm{\forall }i\ge 1$. Then:

• (1)
  $$W_n$$

  is nonexpansive and $F(W_n)={\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$ for each $n\ge 1$;

• (2) for each $x\in C$ and for each positive integer k, the ${lim}_{n\to \mathrm{\infty }}{U}_{n,k}x$ exists;

• (3) the mapping $U:C\to C$ defined by

  $$Ux=\underset{n\to \mathrm{\infty }}{lim}{W}_{n}x=\underset{n\to \mathrm{\infty }}{lim}{U}_{n,1}x,x\in C$$

is a nonexpansive mapping satisfying $F(U)={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_i)$ and it is called the W-mapping generated by $T_1,T_2,\dots$ and ${\lambda }_1,{\lambda }_2,\dots$ ;

• (4)
  $${lim}_{m,n\to \mathrm{\infty }}{sup}_{x\in K}\parallel W_{m}x-W_{n}x\parallel =0$$

  for any bounded subset K of E.

3 Main results

In this section, we introduce our algorithm and prove its strong convergence.

Theorem 3.1 Let C be a closed convex subset of a real Hilbert space H. Let F be a bifunction from $H\times H$ into ℝ satisfying (A1)-(A4). Let f be a contraction of H into itself with $\alpha \in (0,1)$, and let $T_n$ be a sequence of nonexpansive mappings of C into itself such that $\mathrm{\Omega }={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)\cap EP(F)\ne \mathrm{\varnothing }$. Let $A:H\to H$ be a strongly positive bounded linear operator with a coefficient $\overline{\gamma }>0$ with $0<\gamma <\frac{\overline{\gamma }}{\alpha }$. Let ${\lambda }_1,{\lambda }_2,\dots$ be a sequence of real numbers such that $0<{\lambda }_n\le b<1$ for every $n=1,2,\dots$ . Let $W_n$ be a W-mapping of C into itself generated by $T_n,T_{n-1},\dots ,T_1$ and ${\lambda }_n,{\lambda }_{n-1},\dots ,{\lambda }_1$. Let U be defined by $Ux={lim}_{n\to \mathrm{\infty }}{W}_{n}x={lim}_{n\to \mathrm{\infty }}{U}_{n,1}x$ for every $x\in C$. Let $\{x_n\}$ and $\{u_n\}$ be sequences generated by $x_1\in H$ and

$$\{\begin{array}{c}F(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in H,\hfill \\ x_{n+1}=P_C[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n],\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

(3.1)

where $\{{\alpha }_n\}$ is a sequence in $(0,1)$ and $\{r_n\}$ is a sequence in $[0,\mathrm{\infty })$. Suppose that $\{{\alpha }_n\}$ and $\{r_n\}$ satisfy the following conditions:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

(C3) ${lim}_{n\to \mathrm{\infty }}{r}_n=r>0$.

Then both $\{x_n\}$ and $\{u_n\}$ converge strongly to $x^{\ast }\in \mathrm{\Omega }$, which is the unique solution of the variational inequality

$$〈(A-\gamma f)x^{\ast },x^{\ast }-\hat{x}〉\ge 0,\mathrm{\forall }\hat{x}\in \mathrm{\Omega }.$$

(3.2)

Equivalently, one has $x^{\ast }=P_{\mathrm{\Omega }}(I-A+\gamma f)(x^{\ast })$.

Proof We observe that $P_{\mathrm{\Omega }}(\gamma f+(I-A))$ is a contraction. Indeed, for all $x,y\in H$, we have

$$\begin{array}{rcl}\parallel P_{\mathrm{\Omega }}(\gamma f+(I-A))(x)-P_{\mathrm{\Omega }}(\gamma f+(I-A))(y)\parallel & \le & \parallel (\gamma f+(I-A))(x)-(\gamma f+(I-A))(y)\parallel \\ \le & \gamma \parallel f(x)-f(y)\parallel +\parallel I-A\parallel \parallel x-y\parallel \\ \le & \gamma \alpha \parallel x-y\parallel +(1-\overline{\gamma })\parallel x-y\parallel \\ <& (1-(\overline{\gamma }-\gamma \alpha ))\parallel x-y\parallel .\end{array}$$

Banach’s contraction mapping principle guarantees that $P_{\mathrm{\Omega }}(\gamma f+(I-A))$ has a unique fixed point, say $x^{\ast }\in H$. That is, $x^{\ast }=P_{\mathrm{\Omega }}(\gamma f+(I-A))(x^{\ast })$. Note that by Lemma 2.4, we can write

$$x_{n+1}=P_C[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{T}_{r_n}{x}_n],$$

where

$$T_{r_n}(x)=\{z\in H:F(z,y)+\frac{1}{r_n}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in H\}.$$

Moreover, since ${\alpha }_n\to 0$ as $n\to \mathrm{\infty }$ by condition (C1), we assume that ${\alpha }_n\le {\parallel A\parallel }^{-1}$ for all $n\in \mathbb{N}$. From Lemma 2.8, we know that if $0<\rho <{\parallel A\parallel }^{-1}$, then $\parallel I-\rho A\parallel \le 1-\rho \overline{\gamma }$. We divide the proof into seven steps as follows.

Step 1. Show that the sequences $\{x_n\}$ and $\{u_n\}$ are bounded.

Let $\hat{x}\in \mathrm{\Omega }$. Then $\hat{x}\in EP(F)$. From Lemma 2.4, we have

$$\parallel u_n-\hat{x}\parallel =\parallel T_{r_n}{x}_n-T_{r_n}\hat{x}\parallel \le \parallel x_n-\hat{x}\parallel .$$

Thus, we have

$$\begin{array}{rcl}\parallel x_{n+1}-\hat{x}\parallel & =& \parallel P_C[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n]-\hat{x}\parallel \\ \le & \parallel {\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n-\hat{x}\parallel \\ \le & {\alpha }_n\gamma \parallel f(x_n)-f(\hat{x})\parallel +\parallel I-{\alpha }_{n}A\parallel \parallel W_n{u}_n-\hat{x}\parallel +{\alpha }_n\parallel \gamma f(\hat{x})-A\hat{x}\parallel \\ \le & {\alpha }_n\gamma \alpha \parallel x_n-\hat{x}\parallel +(1-{\alpha }_n\overline{\gamma })\parallel u_n-\hat{x}\parallel +{\alpha }_n\parallel \gamma f(\hat{x})-A\hat{x}\parallel \\ =& (1-{\alpha }_n(\overline{\gamma }-\gamma \alpha ))\parallel x_n-\hat{x}\parallel +{\alpha }_n(\overline{\gamma }-\gamma \alpha )\frac{\parallel \gamma f(\hat{x})-A\hat{x}\parallel }{(\overline{\gamma }-\gamma \alpha )}.\end{array}$$

By induction, we have

$$\parallel x_n-\hat{x}\parallel \le max\{\parallel x_1-\hat{x}\parallel ,\frac{\parallel \gamma f(\hat{x})-A\hat{x}\parallel }{\overline{\gamma }-\gamma \alpha }\},\mathrm{\forall }n\ge 0.$$

This shows that the sequence $\{x_n\}$ is bounded, so are $\{u_n\}$, $\{f(x_n)\}$ and $\{W_n{u}_n\}$.

Step 2. Show that $\parallel W_{n+1}{u}_n-W_n{u}_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Let $\hat{x}\in \mathrm{\Omega }$. Since $T_i$ and $U_{n,i}$ are nonexpansive and $T_i\hat{x}=\hat{x}=U_{n,i}\hat{x}$ for every $n\in \mathbb{N}$ and $i\le n+1$, it follows that

$$\begin{array}{rcl}\parallel W_{n+1}{u}_n-W_n{u}_n\parallel & =& \parallel {\lambda }_1{T}_1{U}_{n+1,2}{u}_n-{\lambda }_1{T}_1{U}_{n,2}{u}_n\parallel \\ \le & {\lambda }_1\parallel U_{n+1,2}{u}_n-U_{n,2}{u}_n\parallel \\ =& {\lambda }_1\parallel {\lambda }_2{T}_2{U}_{n+1,3}{u}_n-{\lambda }_2{T}_2{U}_{n,3}{u}_n\parallel \\ \le & {\lambda }_1{\lambda }_2\parallel U_{n+1,3}{u}_n-U_{n,3}{u}_n\parallel \\ ⋮\\ \le & \left(\prod _{i=1}^n{\lambda }_i\right)\parallel U_{n+1,n+1}{u}_n-\hat{x}+\hat{x}-U_{n,n+1}{u}_n\parallel \\ \le & \left(\prod _{i=1}^n{\lambda }_i\right)(\parallel U_{n+1,n+1}{u}_n-\hat{x}+\hat{x}-U_{n,n+1}{u}_n\parallel )\\ \le & 2\left(\prod _{i=1}^n{\lambda }_i\right)\parallel u_n-\hat{x}\parallel .\end{array}$$

Since $\{u_n\}$ is bounded and $0<{\lambda }_n\le b<1$ for any $n\in \mathbb{N}$, the following holds:

$$\underset{n\to \mathrm{\infty }}{lim}\parallel W_{n+1}{u}_n-W_n{u}_n\parallel =0.$$

Step 3. Show that $\parallel x_{n+1}-x_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Setting $S=2P_C-I$, we have S is nonexpansive. Note that $W_n=(1-{\lambda }_1)I+{\lambda }_1{T}_1{U}_{n,2}$. Then we can write

$$\begin{array}{rcl}{x}_{n+1}& =& \frac{I+S}{2}[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n]\\ =& \frac{1-{\alpha }_n}{2}{W}_n{u}_n+\frac{{\alpha }_n}{2}(\gamma f(x_n)-A{W}_n{u}_n+W_n{u}_n)\\ +\frac{1}{2}S[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n]\\ =& \frac{1-{\alpha }_n}{2}[(1-{\lambda }_1)I+{\lambda }_1{T}_1{U}_{n,2}]u_n+\frac{{\alpha }_n}{2}(\gamma f(x_n)-A{W}_n{u}_n+W_n{u}_n)\\ +\frac{1}{2}S[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n]\\ =& \frac{(1-{\lambda }_1)(1-{\alpha }_n)}{2}{u}_n+\frac{{\lambda }_1(1-{\alpha }_n)}{2}{T}_1{U}_{n,2}{u}_n+\frac{{\alpha }_n}{2}(\gamma f(x_n)-A{W}_n{u}_n+W_n{u}_n)\\ +\frac{1}{2}S[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n].\end{array}$$

(3.3)

Note that

$$0<\underset{n\to \mathrm{\infty }}{lim}\frac{(1-{\lambda }_1)(1-{\alpha }_n)}{2}=\frac{1-{\lambda }_1}{2}<1,$$

and

$$\frac{{\lambda }_1(1-{\alpha }_n)}{2}+\frac{1}{2}=\frac{1+{\lambda }_1}{2}-\frac{{\lambda }_1}{2}{\alpha }_n.$$

From (3.3), we have

$$\begin{array}{rcl}{x}_{n+1}& =& [1-(\frac{1+{\lambda }_1}{2}+\frac{1-{\lambda }_1}{2}{\alpha }_n)]u_n+(\frac{1+{\lambda }_1}{2}+\frac{1-{\lambda }_1}{2}{\alpha }_n)\\ \times (\frac{{\lambda }_1(1-{\alpha }_n)}{2}{T}_1{U}_{n,2}{u}_n+\frac{{\alpha }_n}{2}(\gamma f(x_n)-A{W}_n{u}_n+W_n{u}_n)\\ +\frac{1}{2}S[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n])/(\frac{1+{\lambda }_1}{2}+\frac{1-{\lambda }_1}{2}{\alpha }_n)\\ =& [1-(\frac{1+{\lambda }_1}{2}+\frac{1-{\lambda }_1}{2}{\alpha }_n)]u_n+(\frac{1+{\lambda }_1}{2}+\frac{1-{\lambda }_1}{2}{\alpha }_n)y_n,\end{array}$$

(3.4)

where

$$\begin{array}{rcl}{y}_n& =& (\frac{{\lambda }_1(1-{\alpha }_n)}{2}{T}_1{U}_{n,2}{u}_n+\frac{{\alpha }_n}{2}(\gamma f(x_n)-A{W}_n{u}_n+W_n{u}_n)\\ +\frac{1}{2}S[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n])/(\frac{1+{\lambda }_1}{2}+\frac{1-{\lambda }_1}{2}{\alpha }_n)\\ =& ({\lambda }_1(1-{\alpha }_n)T_1{U}_{n,2}{u}_n+{\alpha }_n(\gamma f(x_n)-A{W}_n{u}_n+W_n{u}_n)\\ +S[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n])/(1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n).\end{array}$$

Set $e_n=\gamma f(x_n)-A{W}_n{u}_n+W_n{u}_n$ and $d_n={\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n$ for all n. Then

$$y_n=\frac{{\lambda }_1(1-{\alpha }_n)T_1{U}_{n,2}{u}_n+{\alpha }_n{e}_n+S{d}_n}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n},\mathrm{\forall }n\ge 0.$$

It follows that

$$\begin{array}{rcl}{y}_{n+1}-y_n& =& \frac{{\lambda }_1(1-{\alpha }_{n+1})T_1{U}_{n+1,2}{u}_{n+1}+{\alpha }_{n+1}{e}_{n+1}+S{d}_{n+1}}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}\\ -\frac{{\lambda }_1(1-{\alpha }_n)T_1{U}_{n,2}{u}_n+{\alpha }_n{e}_n+S{d}_n}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}\\ =& \frac{{\lambda }_1(1-{\alpha }_{n+1})}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}(T_1{U}_{n+1,2}{u}_{n+1}-T_1{U}_{n,2}{u}_n)\\ +(\frac{{\lambda }_1(1-{\alpha }_{n+1})}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}-\frac{{\lambda }_1(1-{\alpha }_n)}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n})T_1{U}_{n,2}{u}_n\\ +\frac{{\alpha }_{n+1}{e}_{n+1}}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}-\frac{{\alpha }_n{e}_n}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}\\ +\frac{S{d}_{n+1}-S{d}_n}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}+(\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}-\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n})S{d}_n.\end{array}$$

Thus,

$$\begin{array}{rcl}\parallel y_{n+1}-y_n\parallel & \le & \frac{{\lambda }_1(1-{\alpha }_{n+1})}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}\parallel T_1{U}_{n+1,2}{u}_{n+1}-T_1{U}_{n,2}{u}_n\parallel \\ +|\frac{{\lambda }_1(1-{\alpha }_{n+1})}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}-\frac{{\lambda }_1(1-{\alpha }_n)}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}|\parallel T_1{U}_{n,2}{u}_n\parallel \\ +\frac{{\alpha }_{n+1}}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}\parallel e_{n+1}\parallel +\frac{{\alpha }_n}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}\parallel e_n\parallel \\ +\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}\parallel S{d}_{n+1}-S{d}_n\parallel \\ +|\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}-\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}|\parallel S{d}_n\parallel .\end{array}$$

Since S is nonexpansive, we obtain that

$$\begin{array}{rcl}\parallel S{d}_{n+1}-S{d}_n\parallel & \le & \parallel d_{n+1}-d_n\parallel \\ =& \parallel {\alpha }_{n+1}\gamma f(x_{n+1})+(I-{\alpha }_{n+1}A)W_{n+1}{u}_{n+1}-({\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n)\parallel \\ \le & {\alpha }_{n+1}\parallel \gamma f(x_{n+1})-A{W}_{n+1}{u}_{n+1}\parallel +{\alpha }_n\parallel \gamma f(x_n)-A{W}_n{u}_n)\parallel \\ +\parallel W_{n+1}{u}_{n+1}-W_n{u}_n\parallel \\ \le & {\alpha }_{n+1}\parallel \gamma f(x_{n+1})-A{W}_{n+1}{u}_{n+1}\parallel +{\alpha }_n\parallel \gamma f(x_n)-A{W}_n{u}_n)\parallel \\ +\parallel W_{n+1}{u}_{n+1}-W_{n+1}{u}_n\parallel +\parallel W_{n+1}{u}_n-W_n{u}_n\parallel \\ \le & {\alpha }_{n+1}\parallel \gamma f(x_{n+1})-A{W}_{n+1}{u}_{n+1}\parallel +{\alpha }_n\parallel \gamma f(x_n)-A{W}_n{u}_n)\parallel +\parallel u_{n+1}-u_n\parallel .\\ +\parallel W_{n+1}{u}_n-W_n{u}_n\parallel .\end{array}$$

Since $T_i$ and $U_{n,i}$ are nonexpansive, we have

$$\begin{array}{rcl}\parallel T_1{U}_{n+1,2}{u}_n-T_1{U}_{n,2}{u}_n\parallel & \le & \parallel U_{n+1,2}{u}_n-U_{n,2}{u}_n\parallel \\ =& \parallel {\lambda }_2{T}_2{U}_{n+1,3}{u}_n-{\lambda }_2{T}_2{U}_{n,3}{u}_n\parallel \\ \le & {\lambda }_2\parallel U_{n+1,3}{u}_n-U_{n,3}{u}_n\parallel \\ \le & \cdots \\ \le & {\lambda }_2\cdots {\lambda }_n\parallel U_{n+1,n+1}{u}_n-U_{n,n+1}{u}_n\parallel \\ \le & M\prod _{i=2}^n{\lambda }_i,\end{array}$$

where $M>0$ is a constant such that $\parallel U_{n+1,n+1}{u}_n-U_{n,n+1}{u}_n\parallel \le M$ for all $n\ge 0$. So,

$$\begin{array}{rcl}\parallel T_1{U}_{n+1,2}{u}_{n+1}-T_1{U}_{n,2}{u}_n\parallel & \le & \parallel T_1{U}_{n+1,2}{u}_{n+1}-T_1{U}_{n+1,2}{u}_n\parallel +\parallel T_1{U}_{n+1,2}{u}_n-T_1{U}_{n,2}{u}_n\parallel \\ \le & \parallel u_{n+1}-u_n\parallel +M\prod _{i=2}^n{\lambda }_i.\end{array}$$

Hence,

$$\begin{array}{rcl}\parallel y_{n+1}-y_n\parallel & \le & \frac{{\lambda }_1(1-{\alpha }_{n+1})}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}\parallel u_{n+1}-u_n\parallel +M\prod _{i=2}^n{\lambda }_i\\ +|\frac{{\lambda }_1(1-{\alpha }_{n+1})}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}-\frac{{\lambda }_1(1-{\alpha }_n)}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}|\parallel T_1{U}_{n,2}{u}_n\parallel \\ +\frac{{\alpha }_{n+1}}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}\parallel e_{n+1}\parallel +\frac{{\alpha }_n}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}\parallel e_n\parallel \\ +\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}({\alpha }_{n+1}\parallel \gamma f(x_{n+1})-A{W}_{n+1}{u}_{n+1}\parallel \\ +{\alpha }_n\parallel \gamma f(x_n)-A{W}_n{u}_n\parallel +\parallel u_{n+1}-u_n\parallel )+\parallel W_{n+1}{u}_n-W_n{u}_n\parallel \\ +|\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}-\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}|\parallel S{d}_n\parallel \\ =& \frac{{\lambda }_1(1-{\alpha }_{n+1})}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}\parallel u_{n+1}-u_n\parallel \\ +|\frac{{\lambda }_1(1-{\alpha }_{n+1})}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}-\frac{{\lambda }_1(1-{\alpha }_n)}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}|\parallel T_1{U}_{n,2}{u}_n\parallel \\ +\frac{{\alpha }_{n+1}}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}\parallel e_{n+1}\parallel +\frac{{\alpha }_n}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}\parallel e_n\parallel \\ +\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}({\alpha }_{n+1}\parallel \gamma f(x_{n+1})-A{W}_{n+1}{u}_{n+1}\parallel \\ +{\alpha }_n\parallel \gamma f(x_n)-A{W}_n{u}_n\parallel +\parallel u_{n+1}-u_n\parallel )+\parallel W_{n+1}{u}_n-W_n{u}_n\parallel \\ +|\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}-\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}|\parallel S{d}_n\parallel .\end{array}$$

Note that: (1)

By condition (C1), we have

$$\frac{{\lambda }_1(1-{\alpha }_{n+1})}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}-\frac{{\lambda }_1(1-{\alpha }_n)}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}\to 0$$

and

$$\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_{n+1}}-\frac{1}{1+{\lambda }_1+(1-{\lambda }_1){\alpha }_n}\to 0.$$

• (2)
  $$\parallel W_{n+1}{u}_n-W_n{u}_n\parallel \to 0$$

  as $n\to \mathrm{\infty }$ because of Step 2.

Therefore,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel y_{n+1}-y_n\parallel -\parallel u_{n+1}-u_n\parallel )\le 0.$$

By Lemma 2.2, we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-u_n\parallel =0.$$

Hence, from (3.4), we deduce

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}(\frac{1+{\lambda }_1}{2}+\frac{1-{\lambda }_1}{2}{\alpha }_n)\parallel y_n-u_n\parallel =0.$$

(3.5)

Step 4. Show that $\parallel x_n-W_n{u}_n\parallel \to 0$ as $n\to \mathrm{\infty }$. Indeed, we have

$$\begin{array}{rcl}\parallel x_n-W_n{u}_n\parallel & \le & \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-W_n{u}_n\parallel \\ =& \parallel x_n-x_{n+1}\parallel +\parallel P_C[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n]-W_n{u}_n\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +\parallel {\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n-W_n{u}_n\parallel \\ =& \parallel x_n-x_{n+1}\parallel +\parallel -{\alpha }_{n}A{W}_n{u}_n+{\alpha }_n\gamma f(x_n)\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +{\alpha }_n(\parallel A\parallel \parallel W_n{u}_n\parallel +\gamma \parallel f(x_n)\parallel ).\end{array}$$

Then

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-W_n{u}_n\parallel \le \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x_{n+1}\parallel +{\alpha }_n(\parallel A\parallel \parallel W_n{u}_n\parallel +\gamma \parallel f(x_n)\parallel )=0.$$

(3.6)

Thus, from (3.6), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-W_n{u}_n\parallel =0.$$

Step 5. Show that $\parallel x_n-u_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Let $\hat{x}\in \mathrm{\Omega }$. Since $T_{r_n}$ is firmly nonexpansive, it follows

$$\begin{array}{rcl}{\parallel \hat{x}-u_n\parallel }^2& =& {\parallel T_{r_n}\hat{x}-T_{r_n}{x}_n\parallel }^2\\ \le & 〈T_{r_n}{x}_n-T_{r_n}\hat{x},x_n-\hat{x}〉\\ \le & 〈T_{r_n}{x}_n-\hat{x},x_n-\hat{x}〉\\ =& 〈u_n-\hat{x},x_n-\hat{x}〉\\ =& \frac{1}{2}({\parallel u_n-\hat{x}\parallel }^2+{\parallel x_n-\hat{x}\parallel }^2-{\parallel x_n-u_n\parallel }^2).\end{array}$$

Then

$${\parallel u_n-\hat{x}\parallel }^2\le {\parallel x_n-\hat{x}\parallel }^2-{\parallel x_n-u_n\parallel }^2.$$

Since we have

$$\begin{array}{rcl}{\parallel x_{n+1}-\hat{x}\parallel }^2& =& {\parallel P_C[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n]-P_C\hat{x}\parallel }^2\\ =& {\parallel {\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n-\hat{x}\parallel }^2\\ =& {\parallel (I-{\alpha }_{n}A)(W_n{u}_n-\hat{x})+{\alpha }_n(\gamma f(x_n)-A\hat{x})\parallel }^2\\ \le & {(1-{\alpha }_n\overline{\gamma })}^2{\parallel W_n{u}_n-\hat{x}\parallel }^2+2{\alpha }_n〈\gamma f(x_n)-A\hat{x},x_{n+1}-\hat{x}〉\\ \le & {(1-{\alpha }_n\overline{\gamma })}^2{\parallel u_n-\hat{x}\parallel }^2+2{\alpha }_n\gamma 〈f(x_n)-f(\hat{x}),x_{n+1}-\hat{x}〉\\ +2{\alpha }_n〈\gamma f(\hat{x})-A\hat{x},x_{n+1}-\hat{x}〉\\ \le & {(1-{\alpha }_n\overline{\gamma })}^2({\parallel x_n-\hat{x}\parallel }^2-{\parallel x_n-u_n\parallel }^2)+2{\alpha }_n\gamma \alpha \parallel x_n-\hat{x}\parallel \parallel x_{n+1}-\hat{x}\parallel \\ +2{\alpha }_n\parallel \gamma f(\hat{x})-A\hat{x}\parallel \parallel x_{n+1}-\hat{x}\parallel \\ =& (1-2{\alpha }_n\overline{\gamma }+{({\alpha }_n\overline{\gamma })}^2){\parallel x_n-\hat{x}\parallel }^2-{(1-{\alpha }_n\overline{\gamma })}^2{\parallel x_n-u_n\parallel }^2\\ +2{\alpha }_n\gamma \alpha \parallel x_n-\hat{x}\parallel \parallel x_{n+1}-\hat{x}\parallel +2{\alpha }_n\parallel \gamma f(\hat{x})-A\hat{x}\parallel \parallel x_{n+1}-\hat{x}\parallel \\ \le & {\parallel x_n-\hat{x}\parallel }^2+{\alpha }_n{\overline{\gamma }}^2{\parallel x_n-\hat{x}\parallel }^2-{(1-{\alpha }_n\overline{\gamma })}^2{\parallel x_n-u_n\parallel }^2\\ +2{\alpha }_n\gamma \alpha \parallel x_n-\hat{x}\parallel \parallel x_{n+1}-\hat{x}\parallel +2{\alpha }_n\parallel \gamma f(\hat{x})-A\hat{x}\parallel \parallel x_{n+1}-\hat{x}\parallel ,\end{array}$$

and hence

$$\begin{array}{rcl}{(1-{\alpha }_n\overline{\gamma })}^2{\parallel x_n-u_n\parallel }^2& \le & {\parallel x_n-\hat{x}\parallel }^2-{\parallel x_{n+1}-\hat{x}\parallel }^2+{\alpha }_n{\overline{\gamma }}^2{\parallel x_n-\hat{x}\parallel }^2\\ +2{\alpha }_n\gamma \alpha \parallel x_n-\hat{x}\parallel \parallel x_{n+1}-\hat{x}\parallel +2{\alpha }_n\parallel \gamma f(\hat{x})-A\hat{x}\parallel \parallel x_{n+1}-\hat{x}\parallel \\ \le & \parallel x_n-x_{n+1}\parallel (\parallel x_n-\hat{x}\parallel \parallel x_{n+1}-\hat{x}\parallel )+{\alpha }_n{\overline{\gamma }}^2{\parallel x_n-\hat{x}\parallel }^2\\ +2{\alpha }_n\gamma \alpha \parallel x_n-\hat{x}\parallel \parallel x_{n+1}-\hat{x}\parallel +2{\alpha }_n\parallel \gamma f(\hat{x})-A\hat{x}\parallel \parallel x_{n+1}-\hat{x}\parallel .\end{array}$$

Therefore, we have $\parallel x_n-u_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Step 6. Show that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈\gamma f(x^{\ast })-A{x}^{\ast },x_n-x^{\ast }〉\le 0$.

We can choose a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

$$\underset{i\to \mathrm{\infty }}{lim}〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n_i}-x^{\ast }〉=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_n-x^{\ast }〉.$$

Let

$$A(x_{n_i})=\{x\in H:\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_i}-x\parallel =\underset{y\in H}{inf}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_i}-y\parallel \}$$

be the asymptotic center of $\{x_{n_i}\}$. Since $\{x_{n_i}\}$ is bounded and H is a Hilbert space, it is well known that $A(x_{n_i})$ is a singleton; say $A(x_{n_i})=\{\tilde{x}\}$. Set

$$L=\underset{i\in \mathbb{N}}{sup}\parallel \gamma f(x_{n_i})-A{W}_{n_i}{u}_{n_i}\parallel$$

and for every $x\in H$ define

$$Wx=\underset{i\to \mathrm{\infty }}{lim}{W}_{n_i}x$$

(3.7)

and

$$T_r(x)=\{z\in H:F(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in H\}.$$

Note that

$$\begin{array}{rcl}\parallel x_{n_i}-W\tilde{x}\parallel & \le & \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel x_{n_i+1}-W\tilde{x}\parallel \\ =& \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel P_C[{\alpha }_{n_i}\gamma f(x_{n_i})+(I-{\alpha }_{n_i}A)W_{n_i}{u}_{n_i}]-W\tilde{x}\parallel \\ \le & \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel {\alpha }_{n_i}\gamma f(x_{n_i})+(I-{\alpha }_{n_i}A)W_{n_i}{u}_{n_i}-W\tilde{x}\parallel \\ =& \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel W_{n_i}{u}_{n_i}-W\tilde{x}+{\alpha }_{n_i}(\gamma f(x_{n_i})-A{W}_{n_i}{u}_{n_i})\parallel \\ \le & \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel W_{n_i}{u}_{n_i}-W_{n_i}{x}_{n_i}\parallel +\parallel W_{n_i}{x}_{n_i}-W_{n_i}\tilde{x}\parallel \\ +\parallel W_{n_i}\tilde{x}-W\tilde{x}\parallel +{\alpha }_{n_i}L\\ \le & \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel u_{n_i}-x_{n_i}\parallel +\parallel x_{n_i}-\tilde{x}\parallel +\parallel W_{n_i}\tilde{x}-W\tilde{x}\parallel +{\alpha }_{n_i}L.\end{array}$$

By Steps 1-5, condition (C1) and (3.7), we derive

$$\begin{array}{rcl}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_i}-W\tilde{x}\parallel & \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel u_{n_i}-x_{n_i}\parallel +\parallel x_{n_i}-\tilde{x}\parallel +\parallel W_{n_i}\tilde{x}-W\tilde{x}\parallel \\ \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_i}-\tilde{x}\parallel .\end{array}$$

That is, $W\tilde{x}\in A(x_{n_i})$. Therefore $W\tilde{x}=\tilde{x}$. Next, we show that $\tilde{x}=T_r\tilde{x}$.

Note that for any $x\in H$ and $a,b>0$, we have

$$F(T_{a}x,y)+\frac{1}{a}〈y-T_{a}x,T_{a}x-x〉\ge 0,\mathrm{\forall }y\in H$$

and

$$F(T_{b}x,y)+\frac{1}{b}〈y-T_{b}x,T_{b}x-x〉\ge 0,\mathrm{\forall }y\in H,$$

then

$$F(T_{a}x,T_{b}x)+\frac{1}{a}〈T_{b}x-T_{a}x,T_{a}x-x〉\ge 0$$

and

$$F(T_{b}x,T_{a}x)+\frac{1}{b}〈T_{a}x-T_{b}x,T_{a}x-x〉\ge 0.$$

Summing up the last inequalities and using (A2), we obtain

$$〈T_{a}x-T_{b}x,\frac{T_{b}x-x}{b}-\frac{T_{a}x-x}{a}〉\ge 0.$$

Hence we have

$$\begin{array}{rcl}0& \le & 〈T_{a}x-T_{b}x,T_{b}x-x-\frac{b}{a}(T_{a}x-x)〉\\ =& 〈T_{a}x-T_{b}x,T_{b}x-T_{a}x+T_{a}x-x-\frac{b}{a}(T_{a}x-x)〉\\ =& 〈T_{a}x-T_{b}x,(T_{b}x-T_{a}x)+(1-\frac{b}{a})(T_{a}x-x)〉\\ \le & -{\parallel T_{a}x-T_{b}x\parallel }^2+|1-\frac{b}{a}|\parallel T_{a}x-T_{b}x\parallel (\parallel T_{a}x\parallel +\parallel x\parallel ).\end{array}$$

We derive then

$$\parallel T_{a}x-T_{b}x\parallel \le \frac{|b-a|}{a}(\parallel T_{a}x\parallel +\parallel x\parallel ).$$

It follows that

$$\begin{array}{rcl}\parallel x_{n_i}-T_r\tilde{x}\parallel & \le & \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel x_{n_i+1}-T_r\tilde{x}\parallel \\ =& \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel P_C[{\alpha }_{n_i}\gamma f(x_{n_i})+(I-{\alpha }_{n_i}A)W_{n_i}{u}_{n_i}]-T_r\tilde{x}\parallel \\ \le & \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel {\alpha }_{n_i}\gamma f(x_{n_i})+(I-{\alpha }_{n_i}A)W_{n_i}{u}_{n_i}-T_r\tilde{x}\parallel \\ =& \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel W_{n_i}{u}_{n_i}-T_r\tilde{x}+{\alpha }_{n_i}(\gamma f(x_{n_i})-A{W}_{n_i}{u}_{n_i})\parallel \\ \le & \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel W_{n_i}{u}_{n_i}-x_{n_i}\parallel +\parallel T_{r_{n_i}}{x}_{n_i}-T_{r_{n_i}}\tilde{x}\parallel +\parallel x_{n_i}-T_{r_{n_i}}{x}_{n_i}\parallel \\ +\parallel T_{r_{n_i}}\tilde{x}-T_r\tilde{x}\parallel +{\alpha }_{n_i}L\\ \le & \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel u_{n_i}-x_{n_i}\parallel +\parallel x_{n_i}-\tilde{x}\parallel +\parallel x_{n_i}-u_{n_i}\parallel \\ +\parallel T_{r_{n_i}}\tilde{x}-T_r\tilde{x}\parallel +{\alpha }_{n_i}L\\ \le & \parallel x_{n_i+1}-x_{n_i}\parallel +\parallel u_{n_i}-x_{n_i}\parallel +\parallel x_{n_i}-\tilde{x}\parallel +\parallel x_{n_i}-u_{n_i}\parallel \\ +\frac{|r_{n_i}-r|}{r}(\parallel T_r\tilde{x}\parallel +\parallel \tilde{x}\parallel )+{\alpha }_{n_i}L.\end{array}$$

By Steps 2-5, conditions (C1) and (C3), we obtain

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_i}-T_r\tilde{x}\parallel \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_i}-\tilde{x}\parallel$$

and $\tilde{x}=T_r\tilde{x}$. Thus $\tilde{x}\in F(W)\cap F(T_r)=\mathrm{\Omega }$ by Lemma 2.3 and 2.9. Fix $t\in (0,1)$, $x\in H$ and set $y=\tilde{x}+tx$. Then

$${\parallel x_{n_i}-\tilde{x}-tx\parallel }^2\le {\parallel x_{n_i}-\tilde{x}\parallel }^2+2t〈x,\tilde{x}+tx-x_{n_i}〉.$$

By the minimizing property of $\tilde{x}$ and since ${\parallel \cdot \parallel }^2$ is continuous and increasing in $[0,\mathrm{\infty })$, we have

$$\begin{array}{rcl}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_i}-\tilde{x}\parallel }^2& \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_i}-\tilde{x}-tx\parallel }^2\\ \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_i}-\tilde{x}\parallel }^2+2t\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x,\tilde{x}+tx-x_{n_i}〉.\end{array}$$

Thus,

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x,\tilde{x}+tx-x_{n_i}〉\ge 0.$$

On the other hand,

$$〈x,\tilde{x}-x_{n_i}〉=〈x,\tilde{x}+tx-x_{n_i}〉-t{\parallel x\parallel }^2.$$

Hence we obtain

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x,\tilde{x}-x_{n_i}〉=\underset{t\to 0}{lim}(\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x,\tilde{x}+tx-x_{n_i}〉-t{\parallel x\parallel }^2)\ge 0.$$

Set $x=\gamma f(x^{\ast })-A{x}^{\ast }$. Since $\tilde{x}\in \mathrm{\Omega }$, we obtain

$$\begin{array}{rcl}0& \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },\tilde{x}-x_{n_i}〉\\ \le & 〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },\tilde{x}-x^{\ast }〉+\underset{i\to \mathrm{\infty }}{lim}〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x^{\ast }-x_{n_i}〉\\ \le & \underset{i\to \mathrm{\infty }}{lim}〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x^{\ast }-x_{n_i}〉.\end{array}$$

So that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_n-x^{\ast }〉=-\underset{i\to \mathrm{\infty }}{lim}〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n_i}-x^{\ast }〉\le 0.$$

Step 7. Show that both $\{x_n\}$ and $\{u_n\}$ strongly converge to $x^{\ast }\in \mathrm{\Omega }$, which is the unique solution of the variational inequality (3.2). Indeed, we note that

$${\parallel x_{n+1}-x^{\ast }\parallel }^2=〈x_{n+1}-d_n,x_{n+1}-x^{\ast }〉+〈d_n-x^{\ast },x_{n+1}-x^{\ast }〉.$$

Since $〈x_{n+1}-d_n,x_{n+1}-x^{\ast }〉\le 0$, we get

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & 〈d_n-x^{\ast },x_{n+1}-x^{\ast }〉\\ =& 〈{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{u}_n-x^{\ast },x_{n+1}-x^{\ast }〉\\ =& 〈{\alpha }_n\gamma f(x_n)-{\alpha }_n\gamma f\left(x^{\ast }\right)+W_n{u}_n-{\alpha }_{n}A{W}_n{u}_n-x^{\ast }\\ +{\alpha }_{n}A{x}^{\ast }+{\alpha }_n\gamma f\left(x^{\ast }\right)-{\alpha }_{n}A{x}^{\ast },x_{n+1}-x^{\ast }〉\\ =& 〈{\alpha }_n\gamma (f(x_n)-f\left(x^{\ast }\right))+(I-{\alpha }_{n}A)(W_n{u}_n-x^{\ast }),x_{n+1}-x^{\ast }〉\\ +{\alpha }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n+1}-x^{\ast }〉\\ \le & ({\alpha }_n\gamma \parallel f(x_n)-f\left(x^{\ast }\right)\parallel +\parallel I-{\alpha }_{n}A\parallel \parallel W_n{u}_n-x^{\ast }\parallel )\parallel x_{n+1}-x^{\ast }\parallel \\ +{\alpha }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n+1}-x^{\ast }〉\\ \le & (1-{\alpha }_n(\overline{\gamma }-\gamma \alpha ))\parallel x_n-x^{\ast }\parallel \parallel x_{n+1}-x^{\ast }\parallel +{\alpha }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n+1}-x^{\ast }〉\\ \le & \left(\frac{{[1-{\alpha }_n(\overline{\gamma }-\gamma \alpha )]}^2}{2}\right){\parallel x_n-x^{\ast }\parallel }^2+\frac{1}{2}{\parallel x_{n+1}-x^{\ast }\parallel }^2\\ +{\alpha }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n+1}-x^{\ast }〉.\end{array}$$

It then follows that

$${\parallel x_{n+1}-x^{\ast }\parallel }^2\le [1-{\alpha }_n(\overline{\gamma }-\gamma \alpha )]{\parallel x_n-x^{\ast }\parallel }^2+2{\alpha }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n+1}-x^{\ast }〉.$$

(3.8)

Let $a_n={\parallel x_n-x^{\ast }\parallel }^2$, ${\gamma }_n={\alpha }_n(\overline{\gamma }-\gamma \alpha )$ and ${\delta }_n=2{\alpha }_n〈\gamma f(x^{\ast })-A{x}^{\ast },x_{n+1}-x^{\ast }〉$.

Then, we can write the last inequality as

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\delta }_n.$$

Note that in virtue of condition (C2), ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$. Moreover,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\delta }_n}{{\gamma }_n}=\frac{1}{\overline{\gamma }-\gamma \alpha }\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}2〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n+1}-x^{\ast }〉.$$

By Step 5, we obtain

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\delta }_n}{{\gamma }_n}\le 0.$$

(3.9)

Now, applying Lemma 2.6 to (3.8), we conclude that $x_n\to x^{\ast }$ as $n\to \mathrm{\infty }$. Furthermore, since $\parallel u_n-x^{\ast }\parallel =\parallel T_{r_n}{x}_n-T_{r_n}{x}^{\ast }\parallel \le \parallel x_n-x^{\ast }\parallel$, we then have that $u_n\to x^{\ast }$ as $n\to \mathrm{\infty }$. The proof is now complete. □

Setting $A\equiv I$ and $\gamma =1$ in Theorem 3.1, we have the following result.

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $F:C\times C\to \mathbb{R}$ be an equilibrium bifunction satisfying the conditions:

• (1) F is monotone, that is, $F(x,y)+F(y,x)\le 0$ for all $x,y\in C$;

• (2) for each $x,y,z\in C$, ${lim}_{t\to 0}F(tz+(1-t)x,y)\le F(x,y)$;

• (3) for each $x\in C$, $y\mapsto F(x,y)$ is convex and lower semicontinuous.

Let ${\{T_i\}}_{i=1}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings of C into C such that ${\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)\cap EP(F)\ne \mathrm{\varnothing }$. Suppose $\{{\alpha }_n\}\subset (0,1)$ and $\{r_n\}\subset (0,\mathrm{\infty })$ satisfy the following conditions:

• (1)
  $${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$$

  and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (2)
  $${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$$

  and ${lim}_{n\to \mathrm{\infty }}(r_{n+1}-r_n)=0$.

Let f be a contraction of C into itself, and let $x_0\in H$ be given arbitrarily. Then the sequences $\{x_n\}$ and $\{y_n\}$ generated iteratively by

$$\{\begin{array}{c}F(y_n,x)+\frac{1}{r_n}〈x-y_n,y_n-x_n〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ x_{n+1}={\alpha }_{n}f(x_n)+(1-{\alpha }_n)W_n{y}_n,\hfill \end{array}$$

converge strongly to $x^{\ast }\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)\cap EP(F)$, the unique solution of the minimization problem

$$\underset{x\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)\cap EP(F)}{min}\frac{1}{2}{\parallel x\parallel }^2-h(x),$$

where h is a potential function for f.

Setting $F=0$ in Theorem 3.1, we have the following result.

Corollary 3.3 ([11])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of nonexpansive mappings from C to C such that the common fixed point set $\mathrm{\Omega }={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)\ne \mathrm{\varnothing }$. Let $f:C\to H$ be an α-contraction and $A:H\to H$ be a strongly positive bounded linear operator with a coefficient $\overline{\gamma }>0$. Let γ be a constant such that $0<\gamma \alpha <\overline{\gamma }$. For an arbitrary initial point $x_0$ belonging to C, one defines a sequence ${\{x_n\}}_{n\ge 0}$ iteratively

$$x_{n+1}=P_C[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)W_n{x}_n],\mathrm{\forall }n\ge 0,$$

(3.10)

where $\{{\alpha }_n\}$ is a real sequence in $[0,1]$. Assume that the sequence $\{{\alpha }_n\}$ satisfies the following conditions:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$.

Then the sequence $\{x_n\}$ generated by (3.10) converges in norm to the unique solution $x^{\ast }$, which solves the following variational inequality:

$$x^{\ast }\in \mathrm{\Omega }\mathit{\text{such that}}〈(A-\gamma f)x^{\ast },x^{\ast }-\hat{x}〉\le 0,\mathrm{\forall }\hat{x}\in \mathrm{\Omega }.$$

(3.11)